Where is she vertical asymplotate of this sunction? \( f(x)=\frac{-2 x^{2}+6 x+8}{x^{2}-1} \)

To find the vertical asymptotes of the function \( f(x) = \frac{-2x^2 + 6x + 8}{x^2 - 1} \), you need to identify the values of \( x \) that make the denominator zero, while ensuring they do not also make the numerator zero. The denominator \( x^2 - 1 \) can be factored as \( (x - 1)(x + 1) \). Setting this equal to zero gives you \( x = 1 \) and \( x = -1 \). Since the numerator does not equal zero at these points, both \( x = 1 \) and \( x = -1 \) are vertical asymptotes. Another tip: always double-check if the factors can be canceled out. In this case, no cancellation is present, confirming that the asymptotes are at \( x = 1 \) and \( x = -1 \). Keep practicing with different functions to sharpen your skills in identifying these critical features!